Are affine SDEs invertible?

If we have an a process $X_t$ with values in $\mathbb{R}^{n \times n}$ which solves a linear Stratonovich SDE $$dX_t = A_t X_t dt + B_t X_t \circ dW_t$$ then the inverse of $X_t$ exists and solves $$dZ_t = - Z_t A_t dt - Z_t B_t \circ dW_t$$ It is easy to see that $X_tZ_t =$ Id by using the product rule. My question is this:

If $X$ instead solves the affine SDE $$dX_t = (A_t X_t + a_t) dt + (B_t X_t + b_t) \circ dW_t$$ does it have an inverse?

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@ mathman : I guess you have to (at least formally), calculate $d(1/X_t)$ using multidimensional Itô's lemma and check under what conditions the result holds true. Best regards – TheBridge Jun 15 '12 at 8:39
@TheBridge : What is $1/X_t$ if $X_t$ is a matrix? – mathman Jun 15 '12 at 11:48
even in 1d , with b fixed and non zero this process will always hit zero. When $X_t$ gets small it looks like an ordinary brownian motion. – mike Jun 15 '12 at 12:41
@mike : In 1-d this is a geometric Brownian motion, so does not hit zero. – mathman Jun 15 '12 at 12:58
not when you add a constant to the volatility. that's $b_t$ (little b) that should be non-zero. – mike Jun 15 '12 at 21:15